What is the horizontal asymptote of $y = (x^2-x-6)/(x-2)$ ?

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Answer 1 · 2015-10-22T23:01:04

$y = (x^2-x-6)/(x-2)$

has a vertical asymptote $x=2$

and an oblique asymptote $y = x+1$

It has no horizontal asymptote.

$y = (x^2-x-6)/(x-2)$

$= (x^2-2x+x-2-4)/(x-2)$

$=(x(x-2)+(x-2)-4)/(x-2)$

$=x+1-4/(x-2)$

As $x->+-oo$ , $4/(x-2) -> 0$

So $y = x+1$ is an oblique asymptote.

When $x = 2$ , the denominator $(x-2)$ is zero but the numerator $(x^2-x-6)$ is non-zero.

So $x = 2$ is a vertical asymptote.

What is the horizontal asymptote of y = (x^2-x-6)/(x-2)y = (x^2-x-6)/(x-2)?